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shusha [124]
3 years ago
13

Urgently!

Physics
1 answer:
adelina 88 [10]3 years ago
6 0

Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

1.1m to 0.04125 m

Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

wavelength = 330/300=1.1 m

wavelength = 330/8000 = 0.04125m

note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

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The man fire a 50-g arrow that moves at an unknown speed. It hits and embeds in a 350-g block that slides on an air track. At th
Natasha_Volkova [10]

Answer:

a)  vAix = 80 m/s

b) The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

Explanation:

Arrow embeds into block

Take in the first instance the system to be the arrow + block (isolated). Establish reference coordinate system with the +x axis running horizontally in the direction of the arrow’s motion. The initial state (i) is the arrow travelling with velocity vAix and the final state (f) is the arrow embedded in the block. Now, apply the component form of the Generalized Impulse Momentum Equation to this system:

pAi + pBi + JonA + JonB = pAf + pBf

pAix + pBix + Jx = pAfx + pBfx

mA*vAix + mB*vBix + 0 = (mA + mB)*vfx

0.05*vAix + 0 = (0.05 + 0.35)*vfx

vAix = 8*vfx        (1)

Arrow embeds into block

Now consider the next phase of motion and take as the system the arrow + block + spring. The initial state (i) is the arrow and block travelling with velocity equivalent to the final velocity from equation 1 (final state velocity in first phase becomes initial velocity in next phase);  vix’ = vfx  and the final state (f) is the arrow + block brought to rest and the spring compressed an amount, Δx = 0.1 m. Now, apply the Generalized Work Energy Principle to the system

Ei + W = Ef

Ki + Usi + W = Kf + Usf

0.5*(mA + mB)*vix’² = 0.5*k*Δx²

(0.05 + 0.35)*vfx² = 4000*(0.1)²

vfx = √(40/0.4) = 10 m/s

Substituting above back into equation 1:

vAix = 8* 10 m/s = 80 m/s

Arrow embeds into block

The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

7 0
3 years ago
What is the unit for IMA?
Dahasolnce [82]

It's unitless.


20 characters

7 0
3 years ago
By newton third law of motion, we know that if a rocket ship pushes down on the ground, the ground will push back up on the rock
Furkat [3]

We don't know that at all. The 3rd law says that the REaction is opposite and EQUAL to the action.  We don't know where that "twice as much" comes from.

6 0
4 years ago
Read 2 more answers
What happens if a mid-ocean ridge occurs on land
Gnesinka [82]

Answer:

Despite being such prominent feature on our planet, much of the mid-ocean ridge system remains a mystery. While we have mapped about half of the global mid-ocean ridge in high resolution, less than one percent of the mid-ocean ridge has been explored in detail using submersibles or remotely operated vehicles. so therefore we do not have enough information about them to know what will happen

Explanation:

A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary. Mid-ocean ridges occur along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing enormous volcanic eruptions of basalt. The speed of spreading affects the shape of a ridge  slower spreading rates result in steep, irregular topography while faster spreading rates produce much wider profiles and more gentle slopes.

4 0
3 years ago
A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p
anzhelika [568]

Answer:

The final temperature of the air is T_2= 605 K

Explanation:

We can start by doing an energy balance for the closed system

\Delta KE+\Delta PE+ \Delta U = Q - W

where

\Delta KE = the change in kinetic energy.

\Delta PE = the change in potential energy.

\Delta U = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so \Delta KE = 0 and \Delta PE=0

and our energy balance equation is \Delta U = Q - W

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change \Delta U=W+Q using the energy balance equation.

We convert the interval of time to seconds t = 5 \:min = 300\:s

\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t

\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ

We can use the change in specific internal energy \Delta U = m(u_2-u_1) to find the final temperature of the air.

We are given that T_1=300 \:K and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy u_1

u_1=214.07\:\frac{kJ}{kg}

Next, we will calculate the final specific internal energy u_2

\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1

\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}

u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}

With the value u_2=439.07 \:\frac{kJ}{kg} and the ideal gas tables for air we make a regression between the values u = 434.78 \:\frac{kJ}{kg},T=600 \:K and u = 442.42 \:\frac{kJ}{kg}, T=610 \:K and we find that the final temperature T_2 is 605 K.

3 0
3 years ago
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